| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Find CI width or confidence level |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard formulas: finding the sample proportion from the interval midpoint, then using the interval width to find the z-value and hence the confidence level. It involves routine algebraic manipulation and normal distribution table lookup, but is slightly easier than average as it's a direct application of well-practiced techniques with no conceptual challenges. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.4\) or \(\frac{2}{5}\) or \(\frac{26}{65}\) | B1 [1] | No recovery in (ii) for the B mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{"0.4"} + z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.516\) oe | M1 | or \(\text{"0.4"} - z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.284\) or \(z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.116\) oe |
| \(z = \left(0.116 \times \sqrt{\frac{65}{0.4 \times 0.6}}\right) = 1.909\) | A1 | |
| \((\Phi(\text{'1.909'}) = 0.97(18))\) \(2(\text{'0.97'} - 1)\) | M1 | For fully correct method to find \(\alpha\) from their \(z\) |
| \(\alpha = 94\) | A1 [4] | Allow 94.36 or 94.4 or 94.374 |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.4$ or $\frac{2}{5}$ or $\frac{26}{65}$ | B1 [1] | No recovery in (ii) for the B mark |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{"0.4"} + z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.516$ oe | M1 | or $\text{"0.4"} - z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.284$ or $z \times \sqrt{\frac{0.4 \times 0.6}{65}} = 0.116$ oe |
| $z = \left(0.116 \times \sqrt{\frac{65}{0.4 \times 0.6}}\right) = 1.909$ | A1 | |
| $(\Phi(\text{'1.909'}) = 0.97(18))$ $2(\text{'0.97'} - 1)$ | M1 | For fully correct method to find $\alpha$ from their $z$ |
| $\alpha = 94$ | A1 [4] | Allow 94.36 or 94.4 or 94.374 |
---3 From a random sample of 65 people in a certain town, the proportion who own a bicycle was noted. From this result an $\alpha \%$ confidence interval for the proportion, $p$, of all people in the town who own a bicycle was calculated to be $0.284 < p < 0.516$.\\
(i) Find the proportion of people in the sample who own a bicycle.\\
(ii) Calculate the value of $\alpha$ correct to 2 significant figures.
\hfill \mbox{\textit{CAIE S2 2015 Q3 [5]}}